What is f'(x) if f(x)=cos(5x^4) f'(x)=-sin(20x^3) f'(x)=20x^3sin(5x^4) f'(x)=sin(20x^3)

To find the derivative of the function $f(x) = \cos(5x^4)$, we need to use the chain rule.<br /><br />The chain rule states that if $y = f(u)$ and $u = g(x)$, then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.<br /><br />In this case, let $u = 5x^4$. Then $f(u) = \cos(u)$.<br /><br />Now, we can find the derivatives:<br /><br />$\frac{du}{dx} = \frac{d}{dx}(5x^4) = 20x^3$<br /><br />$\frac{dy}{du} = \frac{d}{du}(\cos(u)) = -\sin(u)$<br /><br />Using the chain rule, we have:<br /><br />$f'(x) = \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = -\sin(u) \cdot 20x^3 = -20x^3\sin(5x^4)$<br /><br />Therefore, the correct answer is $f'(x) = -20x^3\sin(5x^4)$.